# Power System Analysis And Operation Engineering Essay

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Power Flow is designed to simulate transmission grid between the power station and market. In that experiment, only single phase transmission system is tested. There are 2 parts of simulation. The first part only use one generator and to supply the load through one transmission line. In part 2, an additional generator is used to simulate.

We can get a basic understanding of power flow system by recording and analysing the transmission process under different load, transmission line and generating situation. After that, a short conclusion is summarized to discuss about the demand and development direction of power transmission system.

## Introduction

Power flow, sometimes called load flow, is a basic tool for simulating power system, which can be used forÂ analysing power grids. The power flow can successfully investigate the 4 main requirements of normal balanced three phase steady-state conditions. Firstly, it is needed to generate the required power for both the load and loss. Secondly, bus voltage should be close to the rated values. In addition, there may be asked limits of real and reactive power, the generators should operate in the range of those limits. Finally, both the transformers and transmission line should not overload to ensure the stability of power system.

In power system, the condition of transmission line is the key factor that affects the performance of a transmission system. In real system, balanced three phase transmission lines are used to transmit power. In that simulation, or in some other situation, we use single-phase transmission line to analysis the influence of transmission line. That kind of transmission line can be seen as two-port network, due to different length and materials of transmission lines, those lines can be seen as Ï€ circuit and separated as short line, medium line and long-length line.

In power flow system, the voltage magnitude and angle at every bus is computed. The real and reactive power, as well as the losses flow in power system is also computed. Sometimes, the voltage angle difference between different buses, the % load and the thermal rating are also calculated out. Just by changing the parameters of the generators, transmission line and the load, computer can calculated the reliable results which are just the same as real power system.

The software used in that simulation is Power World, the basic circuit analysed is the given "model file", which is a single phase transmission circuit and per-unit power system.

## 3.1 Run PowerWorld simulator and open file 'Example 5-4'

Figure 3.1_1 Oneline diagram (single solution)

Figure 3.1_1 is the simulation result for driving requirement load based on that generator and transmission line. By right clicking left bus (Bus1 [1] ), the result is shown in Figure 3.1_2. Because that the right side of that bus is transmission line, so the load is shown as zero.

Figure 3.1_2 Bus 1 information

Then right clicking right bus (bus 2), a similar result is gotten as Figure 3.1_3. The left side of bus 2 is transmission line, so the generator power is 0. Bus 2 is connected to load directly, so the load power is just 2200MW.

Figure 3.1_3 Bus 2 information

Those results are shown in the following table 1. In addition, the voltage angle difference is calculated out at the same time.

Table 3.1_1 Generator and load information

Generator

difference

Real power

2200.1MW

2200MW

-0.1MW

Reactive power

0.9Mvar

0

-0.9Mvar

voltage

765.00KV

764.94KV

-0.06KV

angle

0deg

-21.395deg

-21.395deg

Form Table 3.1_1, it can be recognised that the reactive power loss is 0.9MW, and the voltage angle difference is -21.395deg. For getting the voltage angle difference by analysing, the transmission line parameters are needed. It is easy to get that the power flows into the system is exactly the same as the generator output, in other words, the real and reactive power flows in the transmission circuit are 2200.1MW and 0.9MV correspondingly (Figure 3.1_6 shows the same result).

Figure 3.1_4 Transmission line information(R and X)

Figure 3.1_4 shows that the resistance value(R) is 0â„¦, and the reactance value(X) is 0.08290â„¦. With those values and the generator, load information, the voltage angle difference can be worked out.

Under that analysis, the direction of voltage (V1) at generator is set as reference direction. Because the output power contains reactive part, the direction of current should be lagging to voltage as shown in Figure 3.1_5. The transmission line has reactance(X), whose direction is perpendicular to the current (I), so it will produce reactive power only (actually cost reactive power from the generator). The voltage at bus 2 (V2) is V1-I2*JX. So the voltage angle of generator should be lagging to V1, as the following figure. The load has real power only, so current should have the same direction as voltage (V2). The voltage angle difference is negative for their phase relation in Figure 3.1_5.

æ-°å›¾ç‰‡(3)

Figure 3.1_5 voltage angle difference

That result shows that the load voltage is lagging to generator voltage. So the angle difference is negative.

When the values of resistance and reactance are changed, both the real and reactive power loss will changed as well. When the demand is constant, the increasing of resistance will increase the real power loss. And the reactance will cost the reactive power generated for the generators. Particularly, when the resistance and reactance is too high, it will cause errors in the transmission circuit for that the power cannot flows to the load at all.

Figure 3.1_6 Power flows between bus 1 and bus 2

Figure 3.1_6 is the power flows between two buses, it shows the real and reactive power flows in the transmission circuit as well. For both bus 1 and bus 2, the % loading is 55.0%.

%load= (real power/power limit)*100% (Power limit A is set to 4000MVA as that case)

The thermal rating is the same as power limit. It is 4000MVA for both Bus 1 and Bus 2.

To do further research on transmission line parameter, two more cases are experimented.

The table below is the results of those experiments.

Table 3.1_2 Experiments on transmission line's resistance and reactance

Operation

Original result

Increase resistance by 0.05â„¦

Increase reactance by 0.02 â„¦

Reactive power loss

0.9Mvar

1582.1Mvar

298.2Mvar

Voltage angle difference

-21.395deg

-37.4deg

-28deg

Voltage at Bus 2

764.94KV

361.407KV

738.891KV

Table 3.1_2 shows that the increasing of resistance and reactance can both cause reactive power loss and voltage angle difference grow. They can also decrease the load voltage. . Particularly, when the resistance or reactance is too high, it will cause errors in the transmission circuit for that the power cannot flows to the load at all. Now an upgraded phase diagram can be drawn out to show their relationship intuitively.

æ-°å›¾ç‰‡(4)

Figure 3.1_7 Phase diagram for normal condition

Form the formula V2=V1-I*(R+JX), the angle and magnitude of load voltage is decided by current, reactance, resistance as well as source voltage.

In that part, the load was increased to 2600MW and 500Mvar. Compared with last case, both the real and reactive power demands are increased. Figure 3.2_1 is the entire circuit.

Figure 3.2_1 Oneline diagram (single solution) for case 2

Figure 3.2_2 Bus 1(Generator information) Figure 3.2_3 Bus 2(Load) information

Figure 3.2_2 and 3.2_3 are the information of generator and load correspondingly. The values of real power, reactive power and voltage are just the same as what are shown in Figure 3.2_1. Table 3.2_1 is a summary of that information.

Table 3.2_1 Generator and Load information

Generator

difference

Real power

2599.9MV

2600MW

-0.1MW

Reactive power

1676.4Mvar

500Mvar

-1176.6Mvar

voltage

765.00KV

598.70KV

-166.3KV

angle

0deg

-21.395deg

-21.395deg

Here is a compare between those 2 cases. It obviously shows out the change of voltage, power loss and power flow.

Table 3.2_2 Comparison between two cases

Case 1: 2200MW and 0Mvar

Case 2: 2600MW and 500Mvar

Difference (trend)

Angle difference

-21.4deg

-33.4deg

Increase (magnitude only) [2]

Voltage at bus 2

764.94KV

598.70KV

Decrease

Reactive power loss

0.9Mvar

1176.6Mvar

Increase

Power flow into the transmission line

2200.1MW

0.9Mvar

2599.9MW

1676.4Mvar

Increase(Far form 500Mvar)

The voltage angle difference changed to -33.4 from -21.4. The reason for that is the current increases at case 2 (Both amplitude and angle difference for the new relationship between generated real and reactive power), and the reactance (X) is still 0.08290. Based on the formula Vr=Vs-I*(JX), when I increases, the voltage angle increases. At the same time and for the same reason, the amplitude of bus 2 voltage drops to 598.70KV. The new phase diagram is Figure 3.2_4. The angle between current (I) and source voltage (V1) increases. The angle between load voltage (V2) and source voltage (V1) increases as well but it is less than the current angle gain.

æ-°å›¾ç‰‡(3)

Figure 3.2_4 Phase diagram when reactive power increases

Another significance difference is the reactive power loss. The reason is that the demanded power contains both reactive power and real power. In that case, the current in the system contains imaginary part. In transmission line, the reactance X will cost those imaginary so there is some reactive power loss. Because of the growing of reactive power loss and demand, the reactive power flows into the circuit becomes larger and equals to the total amount of increased load reactive power and lost reactive power. The real power flows into the transmission line increased 400MW, which is the same as changed in load. There is no resistance in transmission line, so no real power loss in that system.

Figure 3.2_5 Power flow in case 2

The above graph is the power flow in that case. The left one is from the generator to the transmission line. It can be recognized that the power flow into the transmission line is 2599.9MW and 1676.4Mvar.The %loading of that case is 77.3%. The thermal rating=apparent power/%loading= 3093.5/77.3%=4000MVA. The right one is the power flow form the circuit to the load. The power flow through that is 2599.9MW and 499.9Mvar. Its %loading is 66.2%. And the thermal rating is 4000MVA as well.

In real world, the demand of load is keeping changing, to regulate system voltage at load part, a reactive power supply can be added at Bus 2 for supplying the required reactive power. Then the reactive power loss on transmission line will decrease as well as the voltage will be stable to protect power system.

## 3.3 Change line parameters and measure effects

In that new case (case 3), the reactance of transmission line decreases to 0.04764â„¦ from 0.08290â„¦ in case 2. The new circuit is shown in 1.3_1.

Figure 3.3_1 Oneline diagram (single solution) for case3

New Picture New Picture (2)

Figure 3.3_2 Bus 1 (Generator) information Figure 3.3_3 Bus 2 (Load) Information

Recall those dialogue boxes, figure 3.3_2&3 are gotten. With the same analysis method, the following information can be picked out.

Table 3.3_1 Generator and Load information (Case 3)

Generator

difference

Real power

2600MV

2600MW

0MW

Reactive power

417.5Mvar

500Mvar

-82.5Mvar

voltage

765.00KV

729.36KV

-35.64KV

angle

0deg

-15.06deg

-15.06deg

Compare case 3 with case 2, the result is shown in Table 3.3_2. The difference of those two circuits is only reactance(X).

Table 3.3_2 Comparison between to case 2 and case 3

Case 2: X=0.08290â„¦

Case 2: X=0.04764â„¦

Difference (trend)

Angle difference

-33.4deg

-15.1deg

decrease (magnitude only)

Voltage at bus 2

598.70KV

729.356KV

increase

Reactive power loss

1176.6Mvar

82.5Mvar

decrease

Power flow into the transmission line

2599.9MW

1676.4Mvar

2600MW

417.5Mvar

Not change

Decrease

Form the table, it can be find that the angle difference decreases. As discussed in Part 1.2, the voltage difference is decided by current and reactance, the decreasing of reactance decreases the angle difference. Similarly, the magnitude of bus 2 voltage increases for the voltage loss on transmission line drops. Figure 3.3_4 is the new phase diagram, compared with 3.2_4, the line presents for I*JX becomes shorter and the V2 line gets longer.

æ-°å›¾ç‰‡(3)

Figure 3.3_4 Phase diagram for case 3

Another significance changing is reactive power loss on transmission, which drops over 1000Mvar. The series reactance decreases a lot (almost being halved), while the current changes a little (according to the changes in generated power), the reactive power loss in transmission line decreases. Based on the result, the reactive power flows into the transmission line decreases for the balance of energy. The real power flows into the transmission line does not change for the resistance on transmission is still zero.

Figure 3.3_5 is the detailed power flow in that case. The power flows into the circuit is 2600MW and 417.5Mvar(Less than case 2). The %loading at bus 1 is 65.8%, and thermal rating is: 2633.3MVA/65.8%=4000MVA. At bus 2, the power for transmission line to the load is 2600MW and 500Mvar. The %loading is 66.2%, it thermal rating is 4000MVA as well.

New Picture (3)

Figure 3.3_5 Power flow of case 3

To upgrade the network, it is mainly to decrease the reactance on transmission line. There are two ways to realize that. The first one is improve line materials. A purer metal has smaller reactance than normal material. But the cost may be very high. An easier way is to use LC circuit. The reactance is caused by capacitance and inductance, and they have contrary effect. When the reactance of one transmission line is measured out, we can add additional capacitors or inductors to find a balance point to minimize reactance.

## 4.1 -4.3 Add new bus, transmission circuit and generator to

A new bus is added to the system in case 1, the nominal voltage is 765KV, the same as that in bus 1. Then the new bus (bus 2) is connected to bus 2 with a transmission whose resistance is 0 and reactance is 0.04764â„¦. When the transmission circuit is finished, a new generator is connected to bus 3. The output of that generator is 250MW (minimum output is 0 and maximum is 250MW). Because the generator is designed to have a power factor varying from 95 leading to 95lagging, the critical VArs absorbed/injected by the generator is 82Mvar. (Min=-82Mvar and max=82Mvar).

The calculating process is:

P=250MW, pf=95%, S=P/pf=263.16MVA P=(S^2-p^2)^0.5=82Mvar

## 4.4 Solve the new 3-bus network

After setting the new generator, it is regulating the voltage at the new bus to 1pu.

The entire system is shown in the following graph named figure 2_1.

New Picture (4)

Figures 4.4_1 new 3-bus network

New Picture (5)

Figures 4.4_ Power Flow for Bus 1 and 2

New Picture (6)

Figures 4.4_3 Power Flow for Bus 3 and 2

The above two figures are the Powers of the new system. From Figure 4.4_1 to 3, the following information can be gotten.

Table 4.4_1 P, Q and voltage information

P(MW)

Q (Mvar)

V (KV)

Voltage angle(deg)

Bus 1 generator

1950

-128

765

0

Bus 3 generator

250

-57

765

-18.8

2200

0

769.381

-17.395

Table 4.4_2 Power Flow and reactive power loss

Original transmission line

New transmission line

Position

Bus 1

Bus 2

Bus 3

Bus 2

P

1949.9MW to Bus 2

1949.9MW from Bus 1

249.9MW to Bus 2

249.9MW from Bus 3

Q

127.8Mvar From Bus 2

63.3Mvar form Bus 1

57.2Mvar form Bus 2

63.5Mvr to Bus 3

48.9%

48.8%

64.1%

64.5%

Reactive power loss

-191.1Mvar

6.3Mvar

Form Figures 4.4_1 and 3.1_1 (with only one generator), the voltage at Bus 2 increases to 769.381KV from 764.942KV. The voltage angle changes to -18.8° form -21.4 °. And the power supplied form the generator at Bus 1 changes at the same time. The real power (P) reduces to 1949.9MW form 2200MW, which is exactly 250MW that generated by the new generator. The reactive power decreases to -127.8Mvar, which means it receives reactive power from the transmission line. It is 0.9Mvar if the new generator is not added.

When there is only one generator, the reactive power loss on transmission line is 0.9Mvar, for the new system, Qloss=Qloss1+Qloss2= -191.1+6.3=-184.8Mvar. The reason for the decreasing of reactive power loss is that the new add circuit makes the total circuit be capacitive which will produce reactive power. And there is no demand of reactive power at load, so two generators absorb reactive power and a negative power loss will be gotten.

Bus 1 is identified as the swing bus in Power World because that it is the reference for the whole system. The voltage amplitude and angle don't changes when the load condition or transmission line condition change. They are 765KV (1pu) and0 degree all the time.

The %loading of each Bus is shown in Table 4.4_2. Based on the transmission line between Bus 1 and Bus 2, here is a comparison of the changes of %loading.

Table 4.4_3 %loading on transmission line between Bus 1 and 2 in different circuit

Bus 1

Bus 2

2 Buses system

55%

55%

3 Buses system

48.9%

48.8%

From Table 4.4_3, it can be recognised that %loading at bus 1 and bus 2 decrease for that the apparent power (S) required form first generator decreases.

When evaluating the power transmission system, the cost and performance, (or the cost-effective) are needed to be considered. To improve the stability and efficiency of a transmission system, High voltage transmission can be used to decrease the power loss. By upgrading the transmission line material to decrease resistance and reactance on transmission line, the stability that system can will be improved. Other methods, such as changing circuit construction in that experiment can also optimize system. It is easy to improve power system in the Lab, but in real world, the most important things need to be considered are the technique limit and cost. Higher voltage means lower stability, better materials need higher costs. What should engineer to do is to find make a reasonable plan to minimize cost and guarantee the stability of system.

## Change load conditions on new 3-bus network

Clicking on "Play" under 'Run Mode', when simulation is running, increase real power at load to 2600MW. During that process, the real power flows through Bus 1 increases, the reactive power increases as well. The real power flows through Bus 3 does not change and the reactive power at the position increases. When the load real power gets to 2600MW, gradually increase the reactive power at load to 500Mvar. While doing this, the real power at two buses don't change, the reactive power keep increasing until the reactive power of generator at bus 3 reaches 82Mvar. The reactive power at bus 1 goes up all the time.

Figure 4.5_1 is the changed transmission system.

New Picture (7)

Figure 4.5_1 Circuit after changing load

From Figure 4.5_1 the following information can be summarized out.

Table 4.5_1 P, Q, V and voltage angle in new case

Generator at Bus 1

Generator at Bus 3

Real Power (MW)

2350

250

2600

Reactive Power (Mvar)

893

82

500

Voltage (KV)

765

676

669.141

Voltage Angle(deg)

0

-26.5

-24.7

New Picture (2)

Figure 4.5_2 Power Flow at Bus 1 and Bus 2

New Picture (3)

Figure .5_3 Power Flow at Bus 3 and Bus 2

Figure 4.5_2 and 3 are power flow of that case. The power flows into transmission line at bus 1 is 2350MW and 893.1Mvar. The power flows through Bus 2 from the line between Bus 1 and 2 are 2350MW and 426.5Mvar. The power flow at Bus 3 is 250MW and 81.9Mvar. The power flow through Bus 2 from the line between Bus 3 and Bus 2 is 250MW and 73.5Mvar.

During the increasing of real and reactive power demand, the generator at Bus 1 generates more real power and reactive power. The generator at Bus 3 keeps outputting 250MW real power, the reactive power increases until it gets to 82Mvar.

At first, the reactive power generated from two generators increase, when the reactive power of the generator at Bus 3 gets to the maximum value, its output voltage drops down to adjust this requirement.

The increasing output of the generator at Bus 1causes the increase of current for the voltage magnitude and angle at Bus 1 do not change. Then the voltage drop and phase changing of voltage at Bus 2 will increase. So the voltage at the load decreases and the voltage angle increases. The generator at Bus 3 will output the same power after its reactive power reaches 82Mvar. When Bus 2 voltage decreases, the voltage at Bus 3 will decrease to balance the voltage drop on transmission line between Bus 3 and Bus 2.At the same time, the current increases. The voltage angle at Bus 3 also increases during that process for the voltage drop on that transmission increases.

Recheck Figure 4.5_1, the reactive power loss on transmission line between Bus 1 and 2 is 446.4Mvar. It is 8.5Mvar on the line between Bus 3 and Bus 2. So the total reactive loss is Qloss=Qloss1+Qloss2=454.9Mvar. It is easy to find that value in previous case is -184.8Mvar. The reactive power loss increases a lot.

To summarize the loss condition, compare four different circuits. The result is list in the following table.

Table 4.5_2 Reactive power loss under different conditions

1 generator

2 generators

2200MW

0Mvar

0.9Mvar

-184.8Mvar

2600MW

500Mvar

1176.6Mvar

454.9Mvar

Form that table, it can be recognized that adding generator can decrease the reactive power loss in transmission system.

When focusing on electricity market, the electricity company should use suitable size and amount of generators. Because of the property of power energy, the demand of consumers changes all the time and needs to be satisfied immediately in real world. When two generators supply one system at the same time, the small size generator works at low load. When the demand of power increases, the larger one starts to operate. Actually, in power system, basic generator can meet the requirement which can decrease the cost. But extra generators should be prepared for the on-peak demand.

There are there ways to add one more transmission line. They are lines between Bus1&2 (Case A), 2&3(Case B) and 1&3 (Case C). The added line is the same as the existing line between Bus 3&2.

Firstly, adding one more line between Bus 1&2. The result is shown in Figure 4.5_4 (a).

C:\Documents and Settings\alb11122\Local Settings\Temporary Internet Files\Content.Word\New Picture (2).bmp

Figure 4.5_4 (a) Adding line between Bus 1&2. (Case A)

When the load is 2600MW+500Mvar, the new line is overload. Then decrease the real power at load gradually, when it gets to 850MW, the new line works normally as shown in Figure 4.5_4 (b).

C:\Documents and Settings\alb11122\Local Settings\Temporary Internet Files\Content.Word\New Picture (7).bmp

In that case, the total reactive power loss is -799.2Mvar.

After that, try the next case, the new circuit is shown in Figure 4.5_5 which has a new transmission line between Bus 3 and 2.

C:\Documents and Settings\alb11122\Local Settings\Temporary Internet Files\Content.Word\New Picture (8).bmp

Figure 4.5_5 Adding line between Bus 3&2 (Case B)

The new system is similar to the basic circuit. The power loss is 465.7Mvar, a little higher than the system without the new line.

The third method is to add the line between Bus 1 and Bus 3. The following diagram is the simulation result. Two transmission lines are overload.

C:\Documents and Settings\alb11122\Local Settings\Temporary Internet Files\Content.Word\New Picture (4).bmp

Figure 4.5_6 (a) Adding line between Bus 1&3 (Case C)

Gradually decrease the real power at the load. When it drops to 700MW, it gets to full load condition on transmission line between Bus 3 and Bus 2. At that condition, the system becomes stable and the total reactive power losses are -792Mvar.

C:\Documents and Settings\alb11122\Local Settings\Temporary Internet Files\Content.Word\New Picture (6).bmp

Because that Case B doesn't have an obvious influence on that system, only Case A and Case C may meet the requirement. Here is an easy comparison between those two cases.

Table 4.5_3 Comparison of Case A and Case C

Case A

Line between Bus 1&2

Case C

Line between Bus 1&3

Reactive power loss

-799.2Mvar

-792Mvar

850MW

700MW

762KV

760KV